EXPLANATION OF RUTHENIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 3, 2015 Ruthenium is a chemical element with symbol Ru and atomic number 44. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Ruthenium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d75s1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of ruthenium (from (E1 to E3 ) are the following: E1 = 7.36 , E2 = 16.76, and E3 = 28.47 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. ' ' EXPLANATION OF Ε1 = 7.36 eV ' '= -E(5s1) Here the E(5s1) represents the binding energy of 5s1. The charges (-43e) of (1s22s22p63s23p63d10 4s2 4p64d7 ) screen the nuclear charge (+44e) and for a perfect screening we would have ζ = 1. However the electron of 5s1 penetrates the 4d7 and leads to the deformations of electron clouds. Thus ζ > 1. Since 5s1 consists of one electron, we apply the Bohr formula to write E(5s1) = 7.36 eV = - (-13.6057)ζ2/n2 Therefore using n = 5 one writes E1 = 7.36 eV = (13.6057)ζ2 / 52 and we get ζ = 3.68 > 1 . ' ' EXPLANATION OF Ε2 = 16.76 eV -E(4d7) + E(4d6 ) Here the E(4d7) represents the binding energy of the 7 electrons (4d7), while the E(4d6) represents the binding energy of 6 electrons (4d6) .The charges (-36e) of (1s22s22p63s23p63d104s24p6 ) screen the nuclear charge (+44e) and for a perfect screening we would have an effective ζ = 8. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p6 but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 8 . Note that 4d7 consists of two pairs (4 electrons) and of three electrons of parallel spin. Thus for the two pairs of opposite spin we apply my formula of 2008, while for the three electrons of parallel spin we apply the Bohr formula. Under this condition one may write -E(4d7) = -2+ (16.95)ζ - 4.1 / n2 -3(-13.6057)ζ2 /n2 On the other hand, since 4d6 consists of one pair (2 electrons of opposite spin) and of four electrons with parallel spin, we may write E(4d6) = 1+ (16.95)ζ - 4.1 / n2 + 4(-13.6057)ζ2 / n2 Therefore Ε2 = 16.76 eV -E(4d7) + E(4d6 ) = - (16.95)ζ + 4.1) / n2 It is of interest to note that in the sub- shell of 4d the experiments of ionizations showed that 4 < n < 5. (See my paper EXPLANATION OF TECHNETIUM IONIZATIONS). Thus using n = 4 one writes (13.6057) ζ2 - (16.95)ζ - 264 = 0 and solving for ζ we get ζ = 5.07 < 8 Whereas using n = 5 one writes (13.6057)ζ2 - (16.95)ζ - 414.9 = 0 and solving for ζ we get ζ = 6.18 < 8 Note that the two electrons of opposite spin (say the 4dx2 ) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. This situation indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalidrelativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' EXPLANATION OF Ε3 = 28.47 eV -E(4d6) + E(4d5 ) Here the E(4d6) represents the binding energy of the 6 electrons (4d6), while the E(4d5) represents the binding energy of 5 electrons (4d5). As in the case of Ε2 the charges (-36e) of (1s22s22p63s23p63d104s24p6 ) screen the nuclear charge (+44e) and for a perfect screening we would have the same effective ζ = 8. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p6 but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 8 . Note that 4d6 consists of one pair (2 electrons of popposite spin) and of four electrons of parallel spin. Thus for the one pair of opposite spin we apply my formula of 2008, while for the four electrons of parallel spin we apply the Bohr formula. Under this condition one may write -E(4d6) = -+ (16.95)ζ - 4.1 / n2 - 4(-13.6057)ζ2 / n2 On the other hand, since 4d5 consists of five electrons with parallel spin, we may write E(4d5) = + 5(-13.6057)ζ2 / n2 Therefore Ε3 = 28.47 eV -E(4d6) + E(4d5 ) = - (16.95)ζ + 4.1) / n2 As in the case of E1 in the sub- shell of 4d the experiments of ionizations showed that 4 < n < 5. Thususing n = 4 one writes (13.6057)ζ2 - (16.95)ζ - 451.42 = 0 and solving for ζ we get ζ = 6.42 < 8 Whereas using n = 5 one writes (13.6057)ζ2 - (16.95)ζ - 707.65 = 0 and solving for ζ we get ζ = 7.86 < 8 Category:Fundamental physics concepts